This is true for abstract nonsense reasons. If $G$ is a group object in a category $\mathcal{C}$, then $G \times S$ is a group object in the slice category $\mathcal{C}_{/ S}$, and there is a natural bijection between $G$-actions on an object in $\mathcal{C}_{/ S}$ (considered as an object in $\mathcal{C}$) and $(G \times S)$-actions on the same object (considered as an object in $\mathcal{C}_{/ S}$), simply because there is a natural isomorphism $(G \times S) \times_S X \cong G \times X$. Thus, we may assume without loss of generality that $S$ is the terminal object in $\mathcal{C}$. But then $X \times Y$ has an obvious $G$-action if $X$ and $Y$ do: namely, the diagonal action.
More explicitly, suppose we have actions $G \times X \to X$, $G \times Y \to Y$, and $G \times S \to S$ and $G$-equivariant morphisms $X \to S$, $Y \to S$. Then, we can define a $G$-action $G \times (X \times_S Y) \to X \times_S Y$ as the unique morphism such that these equations hold: \begin{align} (G \times (X \times_S Y) \to X \times_S Y \to X) & = (G \times (X \times_S Y) \to G \times X \to X) \\ (G \times (X \times_S Y) \to X \times_S Y \to Y) & = (G \times (X \times_S Y) \to G \times Y \to Y) \end{align}